Name: Student ID: Math 314 (Calculus of Several Variables) Exam 3 May 24 28, Instructions:

Transcription

1 Name: Student ID: Section: Instructor: _00 Scott Glasgow Math 314 (Calculus of Several Variables) RED Eam 3 Ma 4 8, 013 Instructions: For questions which require a written answer, show all our work. Full credit will be given onl if the necessar work is shown, justifing our answer. Simplif our answers. Calculators are not allowed. Tetbooks are not allowed. Notes are not allowed. Should ou need more space than is allotted to answer a question, use the back of the page the problem is on and indicate this fact. Talking about the eam with other students before the graded eam is returned to ou is a violation of the Honor Code.

2 Part I: Multiple Choice Mark the correct answer on the bubble sheet provided. 1. (5 points) Which function has the following level curves? asdfasdasdf asdfsdasdf a) f (, ) b) f (, ) 3 3 c) f (, ) 4 4 d) f (, ) e) f (, ) f) f (, ) Solution: The answer is c), b elimination of the other impossibilities if nothing else: a) gives circles, b) allows for large values of those variables, d) gives straight contours, and e) and f) both give hperbolae.

3 . (5 points) Here are three limits: (i) lim, 0,0 4 lim (ii), 0,0 3 6 (iii) lim, 0,0 Which of these limits eist? a) none of them b) onl (i) c) onl (ii) d) onl (iii) e) onl (i) and (ii) f) onl (i) and (iii) g) onl (ii) and (iii) h) all of them Solution: The correct answer is a): ( ) z( ) z( ) z( ) z( ) z( ) lim : lim lim lim ( ) z ( ) zz ( ) ( ) zz ( ) ( ) z( ) z(0) 0 z(0) z(0) z(0) z(0) z(0) z(0) 0 z(0) z(0) z(0) (1) shows the value of the function depends on how the point 0,0 is approached, hence that limit does not eist. Likewise for the second limit above: 3 ( ) z( ) z(0) lim : lim ( ) z( ) z(0) () Also the third does not eist: ( ) ( ) ( ) ( ) ( ) ( ) lim lim lim ( ) ( ) ( ) ( ) ( ) ( ) (0) 0 (0) (0) (0) (0) (0) (0) (0) 0 (0) 1 (0). (3) 3. (5 points) Consider the fact that z f(, ) dz f (, ) d f (, ) d. (4) This suggests that

4 a) z z0 f( 0, 0) 0 f( 0, 0) 0 the surface z f (, ) at the point 0, 0, 0 b) z z f (, ) f (, ) is the equation for the tangent plane of z, is the equation for the line normal to the tangent plane of the surface z f (, ) c) at the point,, z, z f (, ) f (, ) is the equation for the tangent plane of the surface z f (, ) at the point,, z, d) z f (, ) f (, ) is the equation for the line normal to the tangent plane of the surface z f (, ) at the point 0, 0, 0 e) z z0 f( 0, 0) 0 f( 0, 0) 0 the surface z f (, ) at the point,, z, is the equation for the tangent plane of f) none of the above. Solution: The correct answer is a). z, 0 0 0

5 4. (5 points) Let,, F z be differentiable at Then the level surface given b z, and let F z 0, 0, 0,,,, has a unique tangent plane. Its equation is 0 0 0,, 0,0, F z F z (5) a) F0, 0, z0f0, 0, z0fz 0, 0, z0 z 0, b) Fz0, 0, z0z F0, 0, z0 F 0, 0, z0, c) Fz0, 0, z0z z0 F0, 0, z0 0 F 0, 0, z0 0 d) Fz0, 0, z0zz0 F0, 0, z0 0 F 0, 0, z0 0, e) F z F z F z z z, 0, 0, 0 0 0, 0, 0 0 z 0, 0, 0 0 0, f) none of the above. Solution: e) 5. (5 points) A function z f (, ) is differentiable at a point, a) ab, is in the domain of f, b) ab, is not on the boundar of the domain of f, c) the partial derivatives f and at ab,, d) f is continuous at ab,, e) the partial derivatives f ab, and f, f) none of the above. Solution: c) is a theorem. ab if onl f eist on a small disk centered at ab, and are continuous ab eist,

6 Part II: In the following problems, show all work, and simplif our results. 6. (15 points) Theorem: For ever, we have ( ). (6) Proof: We find that ( ) (7) which clearl can t be negative. Use the Theorem above and the squeeze theorem, etc., to prove that the limit lim, 0,0 3 4 (8) eists, and use those theorems to determine the value of the limit. Solution: We compute that lim lim lim, 0,0 4 4, 0,0, 0, lim lim lim, 0,0 4 5, 0,0 4 5, 0,0 4, 0,0, 0,0, 0, lim 1 lim lim 0 0 0, 4 (9) whence lim, 0,0 4 0 lim 0., 0,0 4 (10)

7 7. (10 points) Write out the Chain Rule for the case in which w f(,, z, t) and each of, z, and t are (differentiable) functions of variable u and v. Specificall, determine the functions Auv, and, Buv in the differential dw A u, v du B u, v dv. (11) Solution: We have f f f f z f t Auv,, u u u z u t u f f f f z f t Bu, v. v v v z v t v (1)

8 8. (10 points) Make the definition gh ( ): f( ah, bh) (13) and compute g (0) in terms of an, all, or some of the following numbers: a, b, f(, ), f (, ), f (, ), f (, ), f (, ), f (, ), f (, ). (14) Solution: We have gh ( ) g(0) f( ah, bh) f(, ) thm g(0) : lim lim : D f (, ) f (, ) (, ) 0 0 ab, a f b h h h h (15) f (, ), f (, ) a, b : f (, ) a, b, using a biggish theorem. Without that particular theorem we could write more fundamentall that gh ( ): f( ah, bh) ah bh g( h) f( ah, bh) f ( ah, bh) h h f ( ah, bh) a f ( ah, bh) b, g(0) f ( a0, b0) a f ( a0, b0) b f (, ) a f (, ) b, (16) using the basic theorem about partial derivatives and composed functions, i.e., using a rather general version of the chain rule.

9 9. (10 points) Assuming the function F F(,, z) has a nonvanishing gradient vector at a point ( 0, 0, z0) on the level surface F(,, z) k F(,, z ), (17) determine an equation giving all points (, z, ) on the plane tangent to that level surface (dictated b (17)) at that point ( 0, 0, z 0). Wh is our equation information free when the gradient vector vanishes at ( 0, 0, z 0)? In what wa does the information free equation actuall make sense? Solution: From (17) we immediatel have that such a tangent plane is given in the small b the differential 0 dk df(,, z) F (,, z) d F (,, z) d F (,, z) dz (18) at an given point (, z, ) on the level surface, or b z 0 F (,, z ) df (,, z ) d F (,, z ) dz (19) z at the specific point ( 0, 0, z0) in question. In the large we have (19) is 0 F (,, z ) F (,, z ) F (,, z ) z z (0) z which is the desired equation. If the gradient vector vanishes at the point in question, then the equation is 0 0, which is satisfied b all points (, z, ), and which makes sense because when the gradient vector vanishes, there is no unique tangent plane, all planes are tangent, and the union of their solution sets is all points in three space.

10 10. (10 points) Consider the plane dictated b the equation 0 F (,, z ) F (,, z ) F (,, z ) z z (1) z Find the line that goes through this plane orthogonall at the point ( 0, 0, z 0). Use an valid format to communicate this line unambiguousl. What assumption do ou have to make so that our proposed equation(s) for a line actuall give a line, not just a point? Solution: A vector normal to the plane is n : F(,, z ) F (,, z ), F (,, z ), F (,, z ) () z whence, the vector equation for such a line is r( t) r tn,, z t F (,, z ), F (,, z ), F (,, z ) z tf (,, z ), tf (,, z ), z tf (,, z ) t z (3) This line degenerates to a point iff F( 0, 0, z0) 0.

11 11. (10 points) B using a Lagrange Multiplier, calculate the maimum value of subject to the side condition that gab (, ): f( a, ) f( b, ) (4) a b 1. (5) Also compute the minimum value of gabsubject (, ) to (5). Assume that Solution: Define f (, ), f (, ) 0,0. What does our computation prove? hab (, ): a b ab, (6) so that constraint (5) is the equation The constrained etrema happen eactl when hab (, ) 1. (7) gab (, ), hab (, ) (8) form a linearl dependent set. (This is equivalent to the actual Lagrange multiplier statement, but more fleible, as one might notice below.) This happens eactl when a certain obvious matri formed from these vectors is singular, i.e., when the following equation holds: T gab (, ) ga( a, b) gb( a, b) f(, ) f(, ) 0 det det det T (, ) ha( a, b) hb( a, b) a b hab f(, ) f(, ) det f(, ) b f(, ) a a b a, b Span f (, ), f (, ) Span f (, ) : f (, ) ab, f(, ), (9) the latter for some to be determined b satisfing the constraint (5). Inserting (9) s last result into (5) gives

12 1 ab, f(, ) f(, ) 1 f(, ) :. (30) For : 1/ f (, ) we get g( a, b): f (, ) a f (, ) bf(, ) a, b f(, ) f(, ) 1 f (, ) f(, ) f(, ), f(, ) (31) while for : 1/ f (, ) we clearl get gab (, ): f(, ) f(, ) 1 f (, ) f(, ) f(, ), f(, ) (3) the former then the maimum value, the latter the minimum value. Noting as in (15) that thm D f (, ) f (, ) (, ), a f b (33) ab we have just proved that the maimum value of the directional derivative at a point is the norm of the gradient vector of the function to be differentiated at said point, the minimum value the negation of that.

13 1. (10 points) Consider taking the second directional derivative as follows: where ab D f (, ): g(0), (34), gh ( ): f( ah, bh). (35) Compute this second directional derivative in terms of an, all, or some of the following numbers: a, b, f(, ), f (, ), f (, ), f (, ), f (, ), f (, ), f (, ). (36) It turns out that ou can write our final result (with the terms in (36)) as ab, M M a D f a b M M b 11 1 (, ), 1 (37) with matri elements M11, M1 M1, M to be determined (b ou). Noting that M M a M a M b M M b M a M b a b a b am a M b bm a M b 1 1 M a M abm bam b , (38) ou see that I ve given ou the hint that our answer should look like h f ( ah, bh) M a M ab M ba M b h0 M a M M abm b M a M abm b (39) If g(0) 0 et g(0) 0, then g has a local min at h 0. If this is true independent of vector ab, with norm 1, then that means that f has a local min at (, ). It turns out that g(0) positive independent of normalized ab, happens iff both of the eigenvalues of the smmetric matri is M11 M1 M11 M1 M : M M M M 1 1 (40)

14 are positive. But the characteristic polnomial for matri M is M M 0det det M M 11 1 I M M M M M M M det M M M M M M (41) And from this it s not hard to show that both eigenvalues are positive if and onl if is positive and so is M11 M det M : M M M (4) 11 1 in which case f has a local minimum at,. But then the criteria for both eigenvalues to be positive (hence get a minimum for f ) is also eactl M M M 0 and M or M 0. (43) To see (43) as equivalent to (4) positive and M11 M positive, note that if M 11 and M are nonpostive, we can t get M11 M positive, and if onl one of them is nonpositive, the product M11M will be nonpositive, and M11M M1 can t be positive. So ultimatel we get that (43) is a sufficient condition for f to have a local min at,. Once ou compute our answer to the first question, which is equivalent to finding the matri elements M11, M1 M1, M, tell me wh it is that ou could have epected that (43) is sufficient for f to have a minimum. Solution: From (16) we alread have g( h) f ( ah, bh) a f ( ah, bh) b ah bh g( h) f ( ah, bh) f ( ah, bh) a h h ah bh f ( ah, bh) f ( ah, bh) b h h f ( ah, bh) a f ( ah, bh) b a f ( ah, bh) a f ( ah, bh) b b g(0) f ( a, ) f ( ba, ) f ( a, ) f ( bb, ) f a f ba f ab f b (, ) (, ) (, ) (, ), (44)

15 where we used the chain rule again, and whence we get D f (, ): f (, ) a f (, ) ba f (, ) ab f (, ) b ab, f ( a, ) f (, ) f (, ) ab f ( b, ) f a f ab f b (, ) (, ) (, ), (45) and whence we can choose M M M f (, ) f (, ) f (, ) f (, ) 11 1 :, M (, ) (, ) (, ) (, ) 1 M f f f f (46) and so the sufficient condition (43) for a minimum is the usual one we ve learned as a theorem. (And we ve done a lot of the work here in proving that theorem.)